Question

A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the ground?B) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the player?

Answer 1

Answer:

4.53482 m/s

4.506 m/s

Explanation:

$m_1$ = Mass of player = 75 kg

$v_1$ = Initial velocity of player = 4.6 m/s

$m_2$ = Mass of ball = 0.47 kg

$v_1$ = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

$(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s$

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

$(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s$

The player's speed is 4.506 m/s